\[\boxed{\mathbf{1127.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\angle AOB = 72{^\circ};\]
\[S_{\text{AOB}} = S.\]
\[\mathbf{Найти:}\]
\[R - ?\]
\[\mathbf{Решение.}\]
\[S_{\text{AOB}} = \frac{\pi R^{2}}{360{^\circ}} \bullet \alpha = \frac{\pi R^{2} \bullet 72{^\circ}}{360{^\circ}} =\]
\[= \frac{\pi R^{2}}{5}\]
\[R^{2} = \frac{5S}{\pi}\]
\[R = \sqrt{\frac{5S}{\pi}}.\]
\[Ответ:R = \sqrt{\frac{5S}{\pi}}.\]